Compact surfaces are classified up to homeomorphism by their fundamental group (defined up to isomorphism). However, 2 compact Riemann surfaces may well be homeomorphic but not conformally equivalent. Still, they are related to their fundamental groups by the fact that the universal cover map is holomorphic : so for say a compact hyperbolic Riemann surface $M$, it can be described as the quotient of the upper half plane under the action of a Fuchsian group which is isomorphic to the fundamental group. That Fuchsian group is not unique though.

My question is this : what is the notion of equivalence needed for two Fuchsian groups to generate the same (hyperbolic) Riemann surface ? It needs a priori to be stronger than just isomorphism.


1 Answer 1


Unless I'm overlooking something:

The groups must be conjugate in $\operatorname{Aut}(\mathbb{H})$.

Say we have two hyperbolic surfaces $S,\, T$ with their universal covers $\pi_S,\, \pi_T$ and a biholomorhism $\varphi \colon S \to T$. Choosing a point $s \in S$ and $\sigma \in \mathbb{H}$ above $s$, and a point $\tau \in \mathbb{H}$ above $t = \varphi(s)$, there is a unique lift of $\varphi$ to $\Phi \colon \mathbb{H} \to \mathbb{H}$ with $\Phi(\sigma) = \tau$ such that the diagram

$$\begin{matrix} \hphantom{XY}\mathbb{H} & \overset{\Phi}{\longrightarrow} & \mathbb{H}\hphantom{XY}\\ \pi_S\downarrow & & \downarrow\pi_T\\ \hphantom{XY}S & \underset{\varphi}{\longrightarrow} & T\hphantom{XY} \end{matrix}$$

is commutative. Now let $\delta$ a deck transformation of $\pi_S$. Then $\pi_T \circ \Phi \circ \delta = \varphi \circ \pi_s \circ \delta = \varphi \circ \pi_S$, hence $\Phi \circ \delta$ is the unique lift of $\varphi$ that maps $\sigma$ to $\Phi(\delta(\sigma))$. Let $\hat{\delta}$ the deck transformation of $\pi_T$ that maps $\tau$ to $\Phi(\delta(\sigma))$. Then we see that $\Phi \circ \delta = \hat{\delta} \circ \Phi$, or $\hat{\delta} = \Phi \circ \delta \circ \Phi^{-1}$. That is, conjugation with $\Phi$ gives an isomorphism $\operatorname{Deck}(\pi_S) \to \operatorname{Deck}(\pi_T)$.

Conversely, if we have $\operatorname{Deck}(\pi_T) = \Phi\operatorname{Deck}(\pi_S) \Phi^{-1}$ for a $\Phi \in \operatorname{Aut}(\mathbb{H})$, we can push down $\Phi$ to a biholomorphism $\varphi$ between $S$ and $T$.

  • $\begingroup$ thanks ! I guess what kind of confused me was that two isomorphic groups need not be conjugated in $\mathrm{Aut}(\mathbb{H})$... It's clearer for me now ! $\endgroup$
    – Khal665
    Jul 12, 2013 at 22:57

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